Đề thi thử THPT Quốc Gia môn Hóa năm 2017 THPT Triệu Sơn 1, Thanh Hoá có lời giải

THPT TRIU SN THI TH THPT QUC GIA LN NM 2017 Mụn: KHOA HC T NHIấN HểA HC ( thi cú 40 cõu / trang) Thi gian lm bi: 50 phỳt, khụng k thi gian phỏt Mó : 132 Cho bit nguyờn t ca cỏc nguyờn t : H =1; C = 12; N = 14; O = 16; Na = 23; Mg = 24; Al = 27; S =32; Cl = 35,5; K = 39; Ca = 40; Cr = 52; Fe = 56; Cu = 64; Zn = 65; Ag = 108; Ba=137 Cõu 1: Khi un núng cht X cú cụng thc phõn t C 5H10O2 vi dung dch NaOH thu c C2H5COONa v ancol Y Y cú tờn l A Ancol Etylic B Ancol Propyolic C Ancol isopropyolic D Ancol Metylic Cõu 2: Cho dung dch lng trng trng tỏc dng vi dung dch axit nitric c,cú hin tng A Kt ta mu tớm B Dung dch mu xanh C Kt ta mu vng D Kt ta mu trng 2+ 2+ + Cõu 3: Cho dóy cỏc cation kim loi :Ca , Cu , Na , Zn2+ Cation kim loi no cú tớnh oxi húa mnh nht dóy A Ca2+ B Cu2+ C Na+ D Zn2+ Cõu 4: Phỏt biu no sau õy l sai ? A Theo chiu tng dn ca in tớch ht nhõn,nhit núng chy ca kim loi kim gim dn B nhit thng, tt c cỏc kim loi kim th u tỏc dng c vi nc C Na2CO3 l nguyờn liu quan trng cụng nghip sn xut thy tinh D Nhụm bn mụi trng khụng khớ v nc l mng oxi Al 2O3 bn vng bo v Cõu 5: Cho s sau : HCl (dư ) CH 3OH, HCl (khan) KOH NaOH,t X1 X2 X3 H2N-CH2COOK X (C4H9O2N) Vy X2 l : A ClH3N-CH2COOH B H2N-CH2-COOH C H2N-CH2-COONa D H2N-CH2COOC2H5 Cõu 6: Cho hn X gm Zn, Fe vo dung dch cha AgNO v Cu(NO3)2, sau phn ng thu c hn hp Y gm kim loi v dung dch Z Cho NaOH d vo dung dch Z thu c kt ta gm hidroxit kim loi.Dung dch Z cha A Zn(NO3)2, AgNO3,Fe(NO3)3 B Zn(NO3)2 ,Fe(NO3)2 C Zn(NO3)2, Fe(NO3)2, Cu(NO3)2 D Zn(NO3)2,Cu(NO3)2, Fe(NO3)3 Cõu 7: Oxit no sau õy l lng tớnh ? A Fe2O3 B CrO C Cr2O3 D CrO3 Cõu 8: in phõn dung dch no sau õy, thỡ cú khớ thoỏt c in cc (ngay t lỳc mi u bt u in phõn) A Cu(NO3)2 B FeCl2 C K2SO4 D FeSO4 Cõu 9: Hp cht H2N-CH2-COOH phn ng c vi : (1) NaOH, (2) HCl, (3) C 2H5OH, (4) HNO2 A (1), (2), (3), (4) B (2), (3), (4) C (1), (2), (4) D (1), (2), (3) Cõu 10: Amin X cú phõn t nh hn 80 Trong phõn t X, nit chim 19,18% v lng.Cho X tỏc dng vi dung dch hn hp gm KNO v HCl thu c ancol Y Oxi húa khụng hon ton Y thu c xeton Z Phỏt biu no sau õy ỳng ? A Tỏch nc Y ch thu c anken nht B.Tờn thay th ca Y l propan-2-ol C.Phõn t X cú mch cacbon khụng phõn nhỏnh D.Trong phõn t X cú liờn kt bi Trang Cõu 11: Dóy kim loi u cú th iu ch bng phng phỏp in phõn dung dch mui ca chỳng l A Na, Cu B Ca, Zn C Fe, Ag D K, Al Cõu 12: Phỏt biu no sau õy khụng ỳng ? A Enzin l nhng cht hu cht cú bn cht protein B Cho glyxin tỏc dng vi HNO cú khớ bay C Phc ng saccarozo cú cụng thc l (C 12H21O11)2Cu D Tetrapeptit thuc loi polipeptit Cõu 13: Cht no sau õy trng thỏi rn iu kin thng ? A Glyxin B Triolein C Etyl aminoaxetat D Anilin Cõu 14: Hũa tan hon ton hn hp X (gm x mol Fe, y mol Cu, z mol Fe 2O3, v t mol Fe3O4) dung dch HCl khụng thy khớ cú khớ bay khi bỡnh, dung dch thu c ch cha mui Mi quan h gia s mol cỏc cht cú hn hp X l : A x + y = 2z + 2t B x + y = z + t C x + y = 2z + 2t D x + y = 2z + 3t Cõu 15: Cho t t tng git ca dung dch cha b mol HCl vo dung dch cha a mol Na 2CO3 thu c V lớt khớ CO2 Ngc li cho t t tng git ca dung dch cha a mol Na 2CO3 vo dung dch cha b mol HCl thu c 2V lớt khớ CO (cỏc th tớch khớ o cựng iu kin) Mi quan h gia a v b l : A a = 0,75b B a = 0,8b C a = 0,35b D a = 0,5b Cõu 16: Dung dch CuSO4 loóng c dựng lm thuc dit nm cho hoa iu ch 800 gam dung dch CuSO4 5%, ngi ta hũa tan CuSO 4.5H2O vo nc Khi lng CuSO 4.5H2O cn dựng l ? A 32,0 gam B 40,0 gam C 62,5 gam D 25,6 gam Cõu 17: Thy phõn 14,6 gam Gly-Ala dung dch NaOH d thu c m gam mui Giỏ tr ca m l : A 20,8 B 18,6 C 22,6 D 20,6 Cõu 18: Thy phõn 44 gam hn hp T gm este cựng cụng thc phõn t C 4H8O2 bng dung dch KOH d Chng ct dung dch sau phn ng thu c hn hp ancol Y v cht rn khan Z un núng Y vi H2SO4 c 1400C, thu c 14,3 gam hn hp cỏc ete Bit cỏc phn ng xy hon ton Khi lng mui Z l A 53,2 gam B 50,0 gam C 34,2 gam D 42,2 gam Cõu 19: Cho hn hp M gm hai cht hu c X, Y Trong ú X l mt axớt hu c hai chc, mch h, khụng phõn nhỏnh (trong phõn t cú mt liờn kt ụi C=C) v Y l ancol no, n chc, mch h t chỏy hon ton 22,32 gam M thu c 14,40 gam H 2O Nu cho 22,32 gam M tỏc dng vi K d thu c 4,256 lớt H2 (ktc) Phn trm lng ca Y M gn nht vi giỏ tr no sau õy? A 27,25% B 62,40% C 72,70% D 37,50% Cõu 20: Cht X cú cụng thc phõn t C 3H4O2, tỏc dng vi dung dch NaOH thu c CHO 2Na Cụng thc cu to ca X l A HCOO-C2H5 B CH3-COOH C CH3-COO-CH3 D HCOO-C2H3 Cõu 21: Dóy gm cỏc cht c sp xp theo chiu tng dn nhit sụi t trỏi sang phi l A HCOOCH3, C2H5OH, HCOOH, CH3COOH B.CH3COOH, HCOOH, C2H5OH, HCOOCH3 C CH3COOH, C2H5OH, HCOOH, HCOOCH3 D.HCOOH, CH3COOH, C2H5OH, HCOOCH3 Cõu 22: Cho vo ng nghim ml dung dch lũng trng trng 10%, thờm tip ml dung dch NaOH 30% v git dung dch CuSO 2% Lc nh ng nghim, hin tng quan sỏt c l A Cú kt ta xanh lam, sau ú tan to dung dch xanh lam B Cú kt ta xanh lam, sau ú kt ta chuyn sang mu gch C Cú kt ta xanh lam, sau ú tan to dung dch mu tớm D Cú kt ta xanh lam, kt ta khụng b tan Cõu 23: Cho cỏc cht sau: HCl, AgNO 3, Cl2, KMnO4/H2SO4 loóng, Cu S cht tỏc dng c vi dung dch Fe(NO3)2 l A B C D Cõu 24: Cho m gam hn hp X gm glyxin v axit glutamic tỏc dng vi 0,4 mol HCl thu c dung Trang dch Y, Y phn ng ti a vi 0,8 mol NaOH thu c 61,9 gam hn hp mui % Khi lng glyxin cú X l A 50,51% B 25,25% C 43,26% D 37,42% Cõu 25: X, Y l hai hp cht hu c n chc phõn t ch cha C, H, O Khi t chỏy X, Y vi s mol bng hoc lng bng u thu c vi t l mol tng ng : v vi t l mol tng ng : S cp cht X, Y tha l A B C D Cõu 26: Polime X dai, bn vi nhit v gi nhit tt nờn dt vi, may qun ỏo m , X l A Poliacrilonitrin B Poli (vinylclorua) C Polibutaien D Polietilen Cõu 27: Cú hn hp, mi hn hp gm cht rn cú s mol bng nhau: Na 2O v Al2O3; Cu v Fe2(SO4)3; KHSO4 v KHCO3; BaCl2 v CuSO4; Fe(NO3)2 v AgNO3 S hn hp cú th tan hon ton nc (d) ch to cỏc cht tan tt nc l A B C D Cõu 28: Cho m gam bt st vo dung dch X cha AgNO3 v Cu(NO3)2 n cỏc phn ng kt thỳc thu c cht rn Y v dung dch Z Cho dung dch Z tỏc dng ht vi dung dch NaOH d, thu c a gam kt ta T gm hai hidroxit kim loi Nung T n lng khụng i thu c b gam cht rn Biu thc liờn h gia m, a, b cú th l A m = 8,225b 7a B m = 8,575b 7a C m = 8,4 3a D m = 9b 6,5a Cõu 29: Thc hin cỏc thớ nghim sau: (1) Nung hn hp Fe v KNO3 khớ tr (2) Cho lung khớ H2 i qua bt CuO nung núng (3) t dõy Mg bỡnh kớn cha y SO2 (4) Nhỳng dõy Ag vo dung dch HNO3 S thớ nghim xy phn ng oxi húa kim loi: A B C D Cõu 30: Hn hp X gm a mol Al v b mol Na Hn hp Y gm b mol Al v a mol Na Thc hin thớ nghim sau Thớ nghim 1: Hũa tan hn hp X vo nc d thu c 5,376 lớt khớ H2, dung dch X1 v m gam cht rn khụng tan Thớ nghim 2: Hũa tan hn hp Y vo nc d thu c dung dch Y1 ú lng NaOH l 1,2 gam Bit th tớch khớ o ktc Tng lng Al hn hp X v Y l A 6,75 gam B 7,02 gam C 7,29 gam D 7,56 gam Cõu 31: Chia dung dch hn hp X gm Al2(SO4)3 v Fe2(SO4)3 thnh hai phn bng Phn mt hũa tan va ỳng 2,56 gam bt Cu Phn hai tỏc dng vi 200 ml dung dch Ba(OH) 1M, thu c 50,5 gam kt ta Cỏc phn ng xy hon ton T l mol gia Al 2(SO4)3 v Fe2(SO4)3 dung dch hn hp X l A : B : C : D : Cõu 32: Nhit phõn mui amoni icromat: (NH4)2Cr2O7 thu c sn phm l A Cr2O3, N2, H2O B Cr2O3, NH3, H2O C CrO3, N2, H2O D CrO3, NH3, H2O Cõu 33: un núng triglyxerit X vi dung dch NaOH va thu c dung dch Y cha mui natri ca axit stearic v oleic em cụ cn dung dch Y thu c 54,84 gam mui Bit X lm mt mu va dung dch cha 0,12 mol Br2 Phn Khi lng phõn t ca X l A 886 B 888 C 884 D 890 Cõu 34: Aminoaxit X (CnH2n+1O2N), ú phn trm lng cacbon chim 51,28% Giỏ tr ca n l A B C D Cõu 35: Cht no sau õy tỏc dng vi tripanmitin A H2 B Dung dch NaOH C Dung dch Br2 D Cu(OH)2 Cõu 36: Cho 5,6 gam Fe vo 200 ml dung dch Cu(NO 3)2 0,5M v HCl 1,2 M thu c khớ NO v m gam kt ta Xỏc nh m Bit rng NO l sn phm kh nht ca NO3- v khụng cú khớ H2 bay A 0,64 B 2,4 C 0,32 D 1,6 Cõu 37: phõn bit cỏc dung dch glucoz, saccaroz v h tinh bt cú th dựng dóy cht no sau õy lm thuc th ? A AgNO3/NH3 v NaOH B Cu(OH) v AgNO3/NH3 Trang C HNO3 v AgNO3/NH3 D Nc brom v NaOH Cõu 38: Cho 35 gam hn hp cỏc amin gm anilin, metylamin, imetylamin, imetylmetylamin tỏc dng va vi 300ml dung dch HCl 1M Khi lng mui khan thu c sau phn ng l: A 45,65 gam B 45,95 gam C 36,095 gam D 56,3 gam Cõu 39: Nhng phn ng húa hc ln lt chng minh rng phõn t glucoz cú nhúm chc CHO v cú nhiu nhúm OH lin k l: A Phn ng gng v phn ng lờn men ru B Phn ng gng v phn ng vi Cu(OH) nhit phũng cho dung dch mu xanh lam C Phn ng to phc vi Cu(OH) v phn ng lờn men ru D Phn ng lờn men ru v phn ng thy phõn Cõu 40: Ly 0,3 mol hn hp X gm H2NC3H5(COOH)2 v H2NCH2COOH cho vo 400 ml dung dch HCl 1M thỡ thu c dung dch Y Y tỏc dng va vi 800 ml dung dch NaOH 1M thu c dung dch Z Lm bay hi Z thu c m gam cht rn khan Giỏ tr ca m l: A 61,9 B 28,8 C 52,2 D 55,2 HT Trang PHN TCH HNG DN GII THI TH THPT TRIU SN - LN Cõu 1: Chn A t C H 5COOC H + NaOH C H 5COONa + C H 5OH etylpropylat Natripropylat etanol Cõu 2: Chn C Khi cho dung dch axit nitric c vo dung dch lũng trng trng thy cú kt ta mu vng xut hin Nhúm R-C6H4-OH ca mt s amino axit protien ó phn ng vi HNO cho hp cht mi mang nhúm NO2 cú mu vng, ng thi protein b ụng t bi HNO3 to thnh kt ta Cõu 3: Chn B Tớnh oxi húa gim dn theo dóy : Cu2+ > Zn2+ > Ca2+ > Na+ Cõu 4: Chn B A ỳng, Theo chiu tng dn in tớch ht nhõn,nhit núng chy ca kim loi kim gim dn B Sai, Mg tỏc dng vi nc nhit cao, Be khụng tỏc dng vi nc mi iu kin nhit C ỳng, Na2CO3 l nguyờn liu quan trng cụng nghip sn xut thy tinh D ỳng, Thc t, cỏc son ni lm bng nhụm cú lp oxit Al 2O3 bn ngn cn khụng cho Al tip xỳc vi H2O iu ny cng xy tng t vi Cr v Zn Cõu 5: Chn A NH2CH2COOC2H5 (X) + NaOH H2N-CH2-COONa (X1) + C2H5OH H2N-CH2-COONa (X1) + HCl ClH3N-CH2-COOH (X2) + NaCl ClH3N-CH2-COOH (X2) + CH3OH ClH3N-CH2-COOCH3 (X3) + H2O ClH3N-CH2-COOCH3 (X3) + 2KOH H2N-CH2-COOK + KCl + CH3OH + H2O Cõu 6: Chn C Ag,Cu :hỗnưhợpưrắnưY NaOH Zn, Fe + AgNO ,Cu(NO ) Zn 2+ , Fe 2+ ,Cu + , NO Fe(OH) ,Cu(OH) + Na ZnO , NaNO 123 44 4 43 4 44 4 43 44 4 43 44 4 43 hỗnưhợpưX dungưdịchưhỗnưhợpư dungưdịchưZ hỗnưhợpưkếtưtủa - Vy dung dch Z chc Zn(NO3)2, Fe(NO3)2 v Cu(NO3)2 Cõu 7: Chn C A Fe2O3 B CrO C Cr2O3 Oxit baz Oxit baz Oxit lng tớnh Cõu 8: Chn C Ti catot H 2O + 2e 2OH - + H2 Ti anot H2O 4H+ + O2 dungưdịchưsảnưphẩm D CrO3 Oxit axit + 4e Bn cht ca in phõn dung dch K 2SO4 l cụ cn dung dch Cõu 9: Chn A H2N-CH2-COOH + NaOH H 2N-CH2-COONa + H2O H2N-CH2-COOH + HCl ClH3N-CH2-COOH + H H2N-CH2-COOH + C2H5OH H2N-CH2-COOC2H5 + H2O 050 C H2N-CH2-COOH + HONO HO -CH2-COOH + N2 + H2O Cõu 11: Chn D Dóy kim loi u cú th iu ch bng phng phỏp in phõn dung dch mui ca chỳng l nhng kim loi hot ng mnh nh kim loi kim, kim th v nhụm Cõu 12: Chn D A ỳng, bn cht ca enzim l nhng cht hu cht cú bn cht protein B ỳng, Cho glyxin tỏc dng vi HNO cú khớ bay Trang 0 C H2N-CH2-COOH + HONO HO -CH2-COOH + N2 + H2O C ỳng, Phc ng saccarozo cú cụng thc l (C 12H21O11)2Cu 2C 12H22O11 + Cu(OH)2 (C12H21O11)2Cu + 2H2O D Sai, peptit c chia thnh hai loi : * Oligopeptit gm cỏc peptit gm cỏc peptit cú t 10 gc aminoaxit * Polipeptit gm cỏc peptit cú t 11 n 50 gc aminoaxit Cõu 13: Chn A - Triolein, Etyl aminoaxetat, Anilin trng thỏi lng Glyxin trng thỏi rn Cõu 14: Chn B BT:e 2n Fe + 2n Cu = 2n Fe3O + 2n Fe 2O x + y = z + t Cõu 15: Chn A - Cho t t a mol HCl vo b mol Na 2CO3 thỡ : n CO2 (1) = n HCl n Na 2CO3 n CO = b a n HCl = 0,5b - Cho t t b mol Na 2CO3 vo a mol HCl thỡ : n CO (2) = n CO2 (1) V b a = = = a = 0,75b - Theo bi ta cú : n CO2 (2) 2V 0,5b Cõu 16: Chn C 800.0,05 = 0,25mol m CuSO4 5H 2O = 62,5(g) - Ta cú: n CuSO = 160 Cõu 17: Chn A t - Phn ng : Gly Ala + 2NaOH GlyNa + AlaNa + H 2O m muối = 97n GlyNa + 111n AlaNa = 20,8(g) Cõu 18: Chn A BTKL - un núng hn hp Y vi H2SO4 thỡ : n H 2O = 0,5n T = 0,25mol m Y = m ete + 18n H 2O = 18,8(g) BTKL - T tỏc dng vi NaOH thỡ m Z = m T + 40n NaOH m Y = 53,2 (g) (với n NaOH = n T = 0,5mol) Cõu 19: Chn D - Khi t 22,32 gam M thỡ : n CO2 = m M 2n H 2O 16n O(trong M) 22,32 2.0,8 16(4n X + n Y ) = 12 12 - p dng bt bóo hũa ta c : 22,32 2.0,8 16(4n X + n Y ) n CO n H 2O = 2n X n Y 0,8 = 2n X n Y 88n X + 4n Y = 11,12(1) 12 - Khi cho lng M trờn tỏc dng vi K d thỡ : 2n X + n Y = 2n H = 0,38(2) - T ta gii h (1) v (2) c : n X = 0,12 molưvàưn Y = 0,14 mol , suy n CO2 = 0,9 mol BT:C - Xột hn hp M ta cú : an X + bn Y = n CO 0,12a + 0,14b = 0,9 a = 4ưvàưb = - Vy X v Y ln lt l : HOOC CH = CH COOH (0,12 mol) v C 3H 7OH (0,14 mol) 0,14.60 100 = 37,63 %m C 3H 7OH(Y) = 22,32 Cõu 20: Chn D t HCOOCH = CH (C 3H O ) + NaOH HCOONa + CH 3CHO Cõu 21: Chn A Cõu 22: Chn C Cõu 23: Chn D Trang Cú cht tỏc dng c vi Fe(NO3)2 l HCl, AgNO3, Cl2 v KMnO4/H2SO4 loóng PT phn ng : 3Fe + + 4H + + NO 3Fe 3+ + NO + 2H 2O Fe(NO3)2 + AgNO3 Fe(NO3)3 + Ag 6Fe(NO3)2 + 3Cl2 4Fe(NO3)3 + 2FeCl3 10Fe(NO3)2 + 2KMnO4 + H2SO4 Fe(NO3)3 + Fe2(SO4)3 + K2SO4 + 2MnSO4 + H2O Cõu 24: Chn A - Xột ton quỏ trỡnh phn ng ta cú h sau : n Gly + 2n Glu = n NaOH n HCl a + 2b = 0, a = 0,2 97n GlyNa + 191n GluNa = m muối 58,5n NaCl 97a + 191b = 38,5 b = 0,1 %m Gly = 75n Gly 100 = 50,51 75n Gly + 147n Glu Cõu 25: Chn B - Khi t chỏy X, Y cú cựng s mol, lng MX = MY n (X) n CO (X) : n CO (Y) = : X : C H 4O n C(X) 2 = = v - Ta cú: n H(Y) n C(Y) Y : C H 8O n H O(X) : n H O(Y) = 1: + Cú ng phõn ca X C2H4O2 l: CH3COOH v HCOOCH3 + Cú S ng phõn ca Y C3H8O l: CH3CH2CH2OH; CH3CH(OH)CH3 v CH3OC2H5 Vy s cp (X, Y) tha l: 3.2 = Cõu 26: Chn A Cõu 27: Chn D Hn hp Na2O v Al2O3: Na2O + H2O 2NaOH + Al2O3 2NaOH 2NaAlO2 + H2O mol mol mol mol - Dung dch sau phn ng cha NaAlO2 l cht tan tt nc Hn hp Cu v Fe2(SO4)3: 2Cu + Fe2(SO4)3 2CuSO4 + FeSO4 mol 0,5 mol - Dung dch sau phn ng cha CuSO4; FeSO4 v Fe2(SO4)3 d l cỏc cht tan tt nc Hn hp KHSO4 v KHCO3: KHSO4 + KHCO3 K2SO4 + CO2 + H2O mol mol - Sau phn ng thu c K 2SO4 tan tt nc nhng khớ CO ớt tan H2O, vy hn hp trờn khụng hon ton tan nc Hn hp BaCl2 v CuSO4: BaCl2 + CuSO4 BaSO4 + CuCl2 mol mol - Sau phn ng thu c BaSO4 kt ta khụng tan nc Hn hp Fe(NO3)2 v AgNO3: Fe(NO3)2 + AgNO3 Fe(NO3)3 + Ag mol mol - Sau phn ng thu c Ag kt ta khụng tan nc Vy cú hn hp cú th tan tt nc d Cõu 28: Chn B - Hng t 1: Fe + 2AgNO3 Fe + Cu(NO3)2 Fe(NO3)2 + 2Ag ; Fe(NO3)2 + Cu Trang mol: x 2x mol 642x7mol 48 4t 48 } Fe + AgNO ,Cu(NO )2 14 43 44 4 43 (x + y) mol m (g) X dungưdịchưX y (Y) Ag : 2x mol, Cu : y mol y x + y mol t y mol 0,5x mol y mol } } 64x7mol48 64y7mol48 78 } NaOH t0 2+ 2+ Fe , Cu , NO Fe(OH)3 ,Cu(OH) Fe O ,CuO 4 4 43 44 4 43 44 43 dungưdịchưZ a (g) b(g) Theo m gam m 56(x + y) = m (1) x+y= Theo a gam 56 (1), (3) 90(x + y) + 98(t y) = a (2) + Ta cú h sau: Theo b gam t y = b m 80(x + y) + 80(t y) = b (3) 80 56 + Thay (x + y) v (t y) vo (2) ta c biu thc: m = 8,575b 7a - Hng t 2: m m 45 mol n Fe 2O3 = mol m Fe(OH) = m gam 56 112 28 10 49b 70m BT: Cu M m Fe2 O3 + m CuO = b m CuO = b m ữgam m Cu(OH) = ữgam 40 45m 49b 70m + = a m = 8,575b 7a - Ta cú: m Fe(OH) + m Cu(OH) = a 28 40 Cõu 29: Chn D - Cỏc phn ng xy ra: BT: Fe n Fe = n Fe(OH) = (1) Nỳng núng KNO3: to 2KNO 2KNO + O sau ú Fe tỏc dng vi O 2: to 3Fe + 2O Fe 3O o t (2) H + CuO Cu + H 2O : phn ng kh oxit kim loi (3) 2Mg + SO 2MgO + S (4) 3Ag + 4HNO3 3AgNO3 + NO + 2H 2O Vy cú phn ng oxi húa kim loi l (1), (3) v (4) Cõu 30: Chn C a mol b mol } } n Al + H 2O Al , Na X1 + H + { dư Ta cú: n Na = b = H = 0,12 mol - Thớ nghim 1: 14 { 43 m gam 0,24 mol hỗnưhợpưX 0,03mol 0,12 mol 64 48 } } + H 2O BT: Na , Na NaOH , NaAlO Khi ú: - Thớ nghim 2: Al a = 0,15 mol 42 43 42 43 0,12 mol a mol dungưdịchưY dungưdịch Y2 m Al = 27.(a + b) = 7, 29 gam Cõu 31: Chn B - Gi x l s mol ca Al2(SO4)3 n Al3+ = 2x mol - Phn 1: hũa tan va ỳng vi 0,04 mol Cu n Cu = n Fe (SO )3 = 0, 04 mol - Phn 2: tỏc dng vi 0,2 mol dung dch Ba(OH)2 thu c kt ta gm: Fe(OH)3: Nhn thy 3n Fe3+ < n OH n Fe(OH)3 = 0, 08 mol m Fe(OH)3 = 8,56 (g) BaSO4: Nu n BaSO = n Ba 2+ = 0, mol m BaSO + m Fe(OH)3 > 50,5 gam n SO 24 = 3(n Al (SO )3 + n Fe (SO )3 ) = 3x + 0,12 m BaSO = 699x + 27,96 (g) Al(OH)3: Xột trng hp to kt ta ca Al(OH)3 (vi n OH cũn li = 0,16 mol) Trang - Trng hp 1: Al(OH)3 khụng b hũa tan + Khi ú 3n Al3+ = 6x n OH = 0,16 x > 0, 0267 m n SO 24 = 3x + 0,12 < 0, x < 0, 0267 (vụ lớ) - Trng hp 2: Al(OH)3 b hũa tan mt phn + Khi ú: n Al(OH)3 = 4n Al3+ n OH = 8x 0,16 m Al(OH)3 = 624x 12, 48 (g) m m Fe(OH)3 + m BaSO + m Al(OH) = 50,85 x = 0, 02 mol Vy n Al (SO )3 = n Fe (SO ) Cõu 32: Chn A o t - Nhit phõn mui amoni icromat: (NH4)2Cr2O7 Cr2O3 + N2 + 4H2O Cõu 33: Chn A - Gi s triglyxerit X c to t gc axit stearic v gc axit oleic Trong phõn t ca X lỳc ny cha liờn kt C=C n Br2 = 0, 06 mol - Khi cho X tỏc dng vi dung dch Br2 thỡ: n X = n C17 H 33COONa = 2n X = 0,12 mol + NaOH - Khi cho X mmui = 54,84 gam (tha yờu cõu bi toỏn) n C17 H 35COONa = n X = 0, 06 mol - Nu nh trng hp trờn khụng tha thi ta suy trng hp cũn li l X c to t gc axit stearic v gc axit oleic Khi ú MX = 888 Vy M X = 886 Cõu 34: Chn C 12n = 0, 5128 n = - Ta cú: %C = 14n + 47 Cõu 35: Chn B - Tripanmitin cú cụng thc (C15H31COO)3C3H5 tham gia phn ng thy phõn mụi trng axit v kim (C15H31COO)3C3H5 + 3NaOH 3C15H31COONa + C3H5(OH)3 Cõu 36: Chn A S oxi húa S kh Fe Fe2+ + 2e 3e + 4H+ + NO3- NO + 2H2O ; Cu2+ + 2e Cu 0,1 0,2 0,18 0,24 0,01 0,1 0,2 (vỡ sau phn ng cú cht rn nờn Fe chuyn lờn Fe2+) 2n Fe 3n NO - Nhn thy: ne nhn > ne cho nCu p = = 0,01 mol mrn = mCu p = 0, 64 (g) Cõu 37: Chn B Thuc th Glucoz Saccaroz H tỡnh bt Cu(OH)2 Phc xanh lam Phc xanh lam Khụng hin tng AgNO3/NH3 Kt ta bc Khụng hin tng Khụng hin tng Cõu 38: Chn B BTKL m muốiưkhan = m amin + 36,5n HCl = 45,95(g) Cõu 39: Chn B Tớnh cht ca ancol a chc (poliancol hay poliol): - Dung dch glucoz hũa tan Cu(OH)2 cho dung dch phc ng glucoz cú mu xanh lam 2C6H12O6 + Cu(OH)2 (C6H11O6)2Cu + 2H2O Tớnh cht ca anehit: - Vi dung dch AgNO3 NH3, un núng (thuc th Tollens) cho phn ng bc Trang o t CH2OH[CHOH]4CHO + 2[Ag(NH3)2]OH CH2OH[CHOH]4COONH4 + 2Ag + 3NH3 + H2O Cõu 40: Chn A n A + n B = n X n A + n B = 0,3 n A = 0,1mol - Xột ton b quỏ trỡnh phn ng ta cú 2n A + n B = n NaOH n HCl 2n A + n B = 0, n B = 0,2 mol - Khi cho dung dch Y tỏc dng vi NaOH thỡ n H 2O = n NaOH = 0,8 mol BTKL m rắnưkhan = 147n A + 75n B + 36,5n HCl + 40n NaOH 18n H O = 61,9 (g) Trang 10